RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQs

In this chapter, we provide RD Sharma Solutions for Chapter 12 Heron’s Formula MCQs for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Chapter 12 Heron’s Formula MCQs Maths pdf, free RD Sharma Solutions for Chapter 12 Heron’s Formula MCQs Maths book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 9
SubjectMaths
ChapterChapter 12
Chapter NameHeron’s Formula
ExerciseMCQs

RD Sharma Solutions for Class 9 Chapter 12 Heron’s Formula MCQs Download PDF

Question 1.
In ∆ABC ≅ ∆LKM, then side of ∆LKM equal to side AC of ∆ABC is
(a) LX
(b) KM
(c) LM
(d) None of these
Solution:
Side AC of ∆ABC = LM of ∆LKM (c)
RD Sharma Class 9 PDF Chapter 12 Heron's Formula

Question 2.
In ∆ABC ≅ ∆ACB, then ∆ABC is isosceles with
(a) AB=AC
(b) AB = BC
(c) AC = BC
(d) None of these
Solution:
∵ ∆ABC ≅ ∆ACB
∴ AB = AC (a)
Heron's Formula Class 9 RD Sharma Solutions

Question 3.
In ∆ABC ≅ ∆PQR, then ∆ABC is congruent to ∆RPQ, then which of the following is not true:
(a) BC = PQ
(b) AC = PR
(c) AB = PQ
(d) QR = BC
Solution:
∵ ∆ABC = ∆PQR
∴ AB = PQ, BC = QR and AC = PR
∴ BC = PQ is not true (a)

Question 4.
In triangles ABC and PQR three equality relations between some parts are as follows: AB = QP, ∠B = ∠P and BC = PR State which of the congruence conditions applies:
(a) SAS
(b) ASA
(c) SSS
(d) RHS
Solution:
In two triangles ∆ABC and ∆PQR,
AB = QP, ∠B = ∠P and BC = PR
The condition apply : SAS (a)
RD Sharma Class 9 Solution Chapter 12 Heron's Formula

Question 5.
In triangles ABC and PQR, if ∠A = ∠R, ∠B = ∠P and AB = RP, then which one of the following congruence conditions applies:
(a) SAS
(b) ASA
(c) SSS
(d) RHS
Solution:
In ∆ABC and ∆PQR,
∠A = ∠R
∠B = ∠P
AB = RP
∴ ∆ABC ≅ ∆PQR (ASA axiom) (b)
Class 9 RD Sharma Solutions Chapter 12 Heron's Formula

Question 6.
If ∆PQR ≅ ∆EFD, then ED =
(a) PQ
(b) QR
(c) PR
(d) None of these
Solution:
∵ ∆PQR = ∆EFD
∴ ED = PR (c)
Class 9 Maths Chapter 12 Heron's Formula RD Sharma Solutions MCQS

Question 7.
If ∆PQR ≅ ∆EFD, then ∠E =
(a) ∠P
(b) ∠Q
(c) ∠R
(d) None of these
Solution:
∵ ∆PQR ≅ ∆EFD
∴ ∠E = ∠P (a)
RD Sharma Book Class 9 PDF Free Download Chapter 12 Heron's Formula

Question 8.
In a ∆ABC, if AB = AC and BC is produced to D such that ∠ACD = 100°, then ∠A =
(a) 20°
(b) 40°
(c) 60°
(d) 80°
Solution:
In ∆ABC, AB = AC
∴ ∠B = ∠C
But Ext. ∠ACD = ∠A + ∠B
RD Sharma Class 9 Book Chapter 12 Heron's Formula
∠ACB + ∠ACD = 180° (Linear pair)
∴ ∠ACB + 100° = 180°
⇒ ∠ACB = 180°-100° = 80°
∴ ∠B = ∠ACD = 80°
But ∠A + ∠B 4- ∠C = 180°
∴ ∠A + 80° + 80° = 180°
⇒∠A+ 160°= 180°
∴ ∠A= 180°- 160° = 20° (a)

Question 9.
In an isosceles triangle, if the vertex angle is twice the sum of the base angles, then the measure of vertex angle of the triangle is
(a) 100°
(b) 120°
(c) 110°
(d) 130°
Solution:
In ∆ABC,
∠A = 2(∠B + ∠C)
= 2∠B + 2∠C
Heron's Formula With Solutions PDF RD Sharma Class 9 Solutions
Adding 2∠A to both sides,
∠A + 2∠A = 2∠A + 2∠B + 2∠C
⇒ 3∠A = 2(∠A + ∠B + ∠C)
⇒ 3∠A = 2 x 180° (∵∠A + ∠B + ∠C = 180° )
⇒ 3∠A = 360°
⇒∠A = 360∘3  = 120°
∴ ∠A = 120° (b)

Question 10.
Which of the following is not a criterion for congruence of triangles?
(a) SAS
(b) SSA
(c) ASA
(d) SSS
Solution:
SSA is not the criterion of congruence of triangles. (b)

Question 11.
In the figure, the measure of ∠B’A’C’ is
(a) 50°
(b) 60°
(c) 70°
(d) 80°
RD Sharma Class 9 Maths Book Questions Chapter 12 Heron's Formula
Solution:
In the figure,
∆ABC ≅ ∆A’B’C’
∴ ∠A = ∠A
⇒3x = 2x- + 20
⇒ 3x – 2x = 20
⇒ x = 20
∠B’A’C’ = 2x + 20 = 2 x 20 + 20
= 40 + 20 = 60° (b)

Question 12.
If ABC and DEF are two triangles such that ∆ABC ≅ ∆FDE and AB = 5 cm, ∠B = 40° and ∠A = 80°. Then, which of the following is true?
(a) DF = 5 cm, ∠F = 60°
(b) DE = 5 cm, ∠E = 60°
(c) DF = 5 cm, ∠E = 60°
(d) DE = 5 cm, ∠D = 40°
Solution:
∵ ∆ABC ≅ ∆FDE,
AB = 5 cm, ∠A = 80°, ∠B = 40°
∴ DF = 5 cm, ∠F = 80°, ∠D = 40°
∴ ∠C =180°- (80° + 40°) = 180° – 120° = 60°
RD Sharma Mathematics Class 9 Solutions Chapter 12 Heron's Formula
∴ ∠E = ∠C = 60°
∴ DF = 5 cm, ∠E = 60° (c)

Question 13.
In the figure, AB ⊥ BE and FE ⊥ BE. If BC = DE and AB = EF, then ∆ABD is congruent to
(a) ∆EFC
(b) ∆ECF
(c) ∆CEF
(d) ∆FEC
Solution Of Rd Sharma Class 9 Chapter 12 Heron's Formula
Solution:
In the figure, AB ⊥ BE, FE ⊥ BE
BC = DE, AB = EF,
then CD + BC = CD + DE BD = CE
In ∆ABD and ∆CEF,
BD = CE (Prove)
AB = FE (Given)
∠B = ∠E (Each 90°)
∴ ∆ABD ≅ ∆FCE (b)

Question 14.
In the figure, if AE || DC and AB = AC, the value of ∠ABD is
(a) 70°
(b) 110°
(c) 120°
(d) 130°
RD Sharma Math Solution Class 9 Chapter 12 Heron's Formula
Solution:
In the figure, AE || DC
∴ ∠1 = 70° (Vertically opposite angles)
∴ ∠1 = ∠2 (Alternate angles)
∠2 = ∠ABC (Base angles of isosceles triangle)
RD Sharma Class 9 Questions Chapter 12 Heron's Formula
∴ ABC = 90°
But ∠ABC + ∠ABD = 180° (Linear pair)
⇒ 70° +∠ABD = 180°
⇒∠ABD = 180°-70°= 110°
∴ ∠ABD =110° (b)

Question 15.
In the figure, ABC is an isosceles triangle whose side AC is produced to E. Through C, CD is drawn parallel to BA. The value of x is
(a) 52°
(b) 76°
(c) 156°
(d) 104°
Maths RD Sharma Class 9 Chapter 12 Heron's Formula
Solution:
In ∆ABC, AB = AC
AC is produced to E
RD Sharma Class 9 Chapter 12 Heron's Formula
CD || BA is drawn
∠ABC = 52°
∴ ∠ACB = 52° (∵ AB = AC)
∴ ∠BAC = 180°-(52° +52°)
= 180°-104° = 76°
∵ AB || CD
∴ ∠ACD = ∠BAC (Alternate angles)
= 76°
and ∠BCE + ∠DCB = 180° (Linear pair)
∠BCE + 52° = 180°
⇒∠BCE = 180°-52°= 128°
∠x + ∠ACD = 380°
⇒ x + 76° = 180°
∴ x= 180°-76°= 104° (d)

Question 16.
In the figure, if AC is bisector of ∠BAD such that AB = 3 cm and AC = 5 cm, then CD =
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 5 cm
RD Sharma Class 9 Solutions Chapter 12 Heron's Formula
Solution:
In the figure, AC is the bisector of ∠BAD, AB = 3 cm, AC = 5 cm
RD Sharma Solutions Class 9 Chapter 12 Heron's Formula
In ∆ABC and ∆ADC,
AC = AC (Common)
∠B = ∠D (Each 90°)
∠BAC = ∠DAC (∵ AC is the bisector of ∠A)
∴ ∆ABC ≅ ∆ADC (AAS axiom)
∴ BC = CD and AB = AD (c.p.c.t.)
Now in right ∆ABC,
AC2 = AB2 + BC2
⇒ (5)2 = (3)2 + BC2
⇒25 = 9 + BC2
⇒ BC2 = 25 – 9 = 16 = (4)2
∴ BC = 4 cm
But CD = BC
∴ CD = 4 cm (c)

Question 17.
D, E, F are the mid-point of the sides BC, CA and AB respectively of ∆ABC. Then ∆DEF is congruent to triangle
(a) ABC
(b) AEF
(c) BFD, CDE
(d) AFE, BFD, CDE
Solution:
In ∆ABC, D, E, F are the mid-points of the sides BC, CA, AB respectively
DE, EF and FD are joined
RD Sharma Class 9 PDF Chapter 12 Heron's Formula
∵ E and F are the mid-points
AC and AB,
∴ EF = 12 BC and EF || BC
Similarly,
DE = 12 AB and DE || AB
DF = 12 AC and DF || AC
∴ ∆DEF is congruent to each of the triangles so formed
∴ ∆DEF is congruent to triangle AFE, BFD, CDE (d)

Question 18.
ABC is an isosceles triangle such that AB = AC and AD is the median to base BC. Then, ∠BAD =
(a) 55°
(b) 70°
(c) 35°
(d) 110°
Heron's Formula Class 9 RD Sharma Solutions
Solution:
In ∆ABC, AB = AC
AD is median to BC
RD Sharma Class 9 Solution Chapter 12 Heron's Formula
∴ BD = DC
In ∆ADB, ∠D = 90°, ∠B = 35°
But ∠B + BAD + ∠D = 180° (Sum of angles of a triangle)
⇒ 35° + ∠BAD + 90° = 180°
⇒∠BAD + 125°= 180°
⇒ ∠BAD = 180°- 125°
⇒∠BAD = 55° (a)

Question 19.
In the figure, X is a point in the interior of square ABCD. AXYZ is also a square. If DY = 3 cm and AZ = 2 cm, then BY =
(a) 5 cm
(b) 6 cm
(c) 7 cm
(d) 8 cm
Class 9 RD Sharma Solutions Chapter 12 Heron's Formula
Solution:
In the figure, ABCD and AXYZ are squares
DY = 3 cm, AZ = 2 cm
Class 9 Maths Chapter 12 Heron's Formula RD Sharma Solutions
DZ = DY + YZ
= DY + Z = 3 + 2 = 5 cm
In ∆ADZ, ∠2 = 90°
AD2 + AZ2 + DZ2 = 22 + 52 cm
= 4 + 25 = 29
In ∠ABX, ∠X = 90°
AB2 = AX2 + BX2
AD2 = AZ2 + BX2
(∵ AB = AD, AX = AZ sides of square)
29 = 22 + BX2
⇒ 29 = 4 + BX2
⇒ BX2 = 29 – 4 = 25 = (5)2
∴ BX = 5 cm (a)

Question 20.
In the figure, ABC is a triangle in which ∠B = 2∠C. D is a point on side BC such that AD bisects ∠BAC and AB = CD. BE is the bisector of ∠B. The measure of ∠BAC is
(a) 72°
(b) 73°
(c) 74°
(d) 95°
RD Sharma Book Class 9 PDF Free Download Chapter 12 Heron's Formula
Solution:
In the figure, ∠B = 2∠C, AD and BE are the bisectors of ∠A and ∠B respectively,
AB = CD
RD Sharma Class 9 Book Chapter 12 Heron's Formula
Heron's Formula With Solutions PDF RD Sharma Class 9 Solutions
RD Sharma Class 9 Maths Book Questions Chapter 12 Heron's Formula

All Chapter RD Sharma Solutions For Class 9 Maths

—————————————————————————–

All Subject NCERT Exemplar Problems Solutions For Class 9

All Subject NCERT Solutions For Class 9

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.

Leave a Comment

Your email address will not be published.